A:简单的异或
https://ac.nowcoder.com/acm/contest/63602/A
题意：

^:相同是0，不同是1

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e5 + 10;
int n, q;
int a[N][40], sum[N][40];

signed main()
{
    cin >> n;
    for(int i = 1;  i <= n; i ++) 
    {
        cin >> a[i][0];
        for(int j = 1; j <= 31; j ++)
            a[i][j] = a[i][0] >> (j - 1) & 1;
        //存第i个数的二进制数的第j位是不是1
    }

    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= 31; j ++)
            sum[i][j] = sum[i - 1][j] + a[i][j];
    //前缀和

    cin >> q;
    int ans;
    while(q --)
    {
        int x = 0, l, r;
        cin >> l >> r;
        for(int i = 1; i <= 31; i ++)
        {
            ans = sum[r][i] - sum[l - 1][i];
            //ans是在l到r范围内第i位的1的个数
            if(ans < r - l + 1 - ans) 
                x += 1 << (i - 1);
            //如果1的个数小于0的个数
            //也就是说：x应该是1的个数大于0的个数时
            //这一位给他存上1 
        }
        cout << x << endl;
    }
    return 0;
}

H:左右横跳
https://ac.nowcoder.com/acm/contest/63602/H?&headNav=acm
题意：

#include <bits/stdc++.h>
using namespace std;
#define int long long

const int N = 1e5 + 10;
int n, m, k;
int a[N], f[N];

signed main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);//这里必须要加速
    cin >> n >> m >> k;
    for(int i = 1; i <= n; i ++)
    {
        for(int j = 0; j < m; j ++)
        {
            int x; cin >> x;
            a[i] = max(a[i], x);//找到每一组的最大值
        }
    }

    f[1] = a[1];
    for(int i = 2; i <= n; i ++)
    {
        f[i] = f[i - 1];//初始化，必须在这里面进行这一步
        if(i - k > 0)
            f[i] = max(f[i], f[i - k] + a[i]);
    }
    cout << f[n] << endl;
    return 0;
}

J：线代高手
https://ac.nowcoder.com/acm/contest/63602/J
题意：


所以就是暴力判断左上角和右下角就行

#include <bits/stdc++.h>
using namespace std;

int n, m;
int a[55], b[55];

int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i ++) cin >> a[i];
    for(int i = 1; i <= m; i ++) cin >> b[i];

    vector<int>aa(n + 1, 0), bb(m + 1, 0);
    for(int i = 1; i <= n; i ++) 
        aa[i] = aa[i - 1] + a[i];
    for(int i = 1; i <= m; i ++) 
        bb[i] = bb[i - 1] + b[i];

    int x; cin >> x;
    int maxn = 0;
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++)
            for(int u = 1; u <= n; u ++)
                for(int o = 1; o <= m; o ++)
                {
                    int sa = aa[u] - aa[i - 1];
                    int sb = bb[o] - bb[j - 1];
                    int s = sa * sb;

                    if(s <= x)
                    {
                        int area = (u - i + 1) * (o - j + 1);
                        maxn = max(area, maxn);
                    }
                }
    cout << maxn << endl;
    return 0;
}